Integer solutions to $x^3=y^3+2y+1$? 
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$

My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
 A: Or you can write $x=y+z$ for some $z\in Z$ (that is $z=x-y$). So we get a quadratic equation on $y$:
$$ 3y^2z+y(3z^2-2)+z^3-1=0$$
which has a discriminant a perfect square: $$d^2 = (3z^2-2)^2-12z(z^3-1)=-3z^4-12z^2+12z+4$$
So we have $$3z^4+12z^2\leq 12z+4$$
and this can not be true for a lot of integers $z$ (in fact only for $0$ and $1$)...
A: Hint: if $y>0$, then $y^3< y^3+2y+1< (y+1)^3$, so the RHS expression cannot be a perfect cube. A similar idea works if $y$ is a small enough negative number, but some negative numbers close to $0$ (or indeed $0$ itself) can provide a solution. 
Try to find a lower bound, and then check the remaining possible values.
A: Solution
Assume that $$x=y+t,~~~t \in \mathbb{Z}.$$
Then $$(y+t)^3=y^3+2y+1,$$ namely $$3ty^2+(3t^2-2)y+t^3-1=0.\tag1$$
If $t=0$, then $y=\dfrac{1}{2}$, which is absurd. Therefore, $t \neq 0$, which implies that $(1)$ could be seen as  a quadratic equation with respect to $y$.
Consider the discriminant for $(1)$. $$\Delta=(3t^2-2)^2-4\cdot 3t \cdot (t^3-1)=-3(t^2+2)^2+12t+16.$$
If $|t|\geq 2$, then $$\Delta=-3(t^2+2)^2+12t+16<-3(t+2)^2+12t+16=4-3t^2<0.$$ Thus, $(1)$ has no real root.
Therefore, the possible values of $t$ are $t=\pm 1.$ Now, we may verify that $t=1$ is the only one solution. Under this case, $$x=1,y=0.$$
A: There are already a lot of solutions but I want to point out that your factorization can be completed to a proof:
Assume that y is nonnegative. From 
$$x^3=y^3+2y+1$$
follows that
$$x^3>y^3$$
and therefore
$$x>y.$$
This implies
$$3y\le 3y^2\le 1 (y^2+y^2+y^2)\lt (x-y)(x^2 + xy + y^2)=2y+1$$
and further
$$y\lt 1 .$$
For negative $y$ a similar argument holds.
A: we first consider only $x> y\ge0$
suppose $x=y+a$
then the equation is $(y+a)^3=y^3+2y+1$
$y^3+3ay^2+3a^2y+a=y^3+2y+1$
$3ay^2+(3a^2-2)y+(a-1)=0$
however, as $a\ge1$, $3a^2-2\ge0$,$a-1\ge0$
If $y>0$, then $3ay^2+(3a^2-2)y+(a-1)>0$, so $y=0$
then, $x^3=1$
$(x,y)=(1,0)$ are only integral solution
A: We have 
$$(y^3+2y+1)-(y-1)^3=3y^2-y+2=\frac{(6y-1)^2+23}{12}>0$$
and
$$(y+2)^3-(y^3+2y+1)=6y^2+10y+7=\frac{(6y+5)^2+17}{6}>0\,.$$
That is,
$$(y-1)^3<y^3+2y+1<(y+2)^3$$
for all $y\in\mathbb{Z}$.  If $y^3+2y+1$ is a cube, then either $y^3+2y+1=y^3$ or $y^3+2y+1=(y+1)^3$, which gives $y=-\dfrac12$, $y=-\dfrac13$, or $y=0$.  That is, $(x,y)=(1,0)$ is the only integer solution to $x^3=y^3+2y+1$.
A: $$x^3=y^3+2y+1\tag{1}$$
We are after integer solutions of $(1)$. On setting $y=x-a$, where $a\in\mathbb{Z}$, we have from $(1)$
$$x^3=(x-a)^3+2(x-a)+1=x^3-3a x^2+(3a^2+2)x-a^3-2a+1\tag{2}$$
which rearranges to
$$3a x^2-(3a^2+2)x+a^3+2a-1=0\tag{3}$$
Solve for $x$
\begin{align*}
x&=\frac{(3a^2+2)\pm\sqrt{(3a^2+2)^2-12a(a^3+2a-1)}}{6a}\\
&=\frac{(3a^2+2)\pm\sqrt{-3a^4-12a^2+12a+4}}{6a}\tag{4}
\end{align*}
The discriminant is
\begin{align*}
-3a^4-12a^2+12a+4&=-3a^4-4(3a^2-3a-1)\\
&=-3a^4-4[3\left((a-\tfrac12)^2-\tfrac7{12}\right)]\\
&=-3a^4-12(a-\tfrac12)^2+7\\
&=7-(3a^4+3(2a-1)^2)
\end{align*}
Now check various integral $a$ for solutions: $a=0$ gives division by zero in $(4)$; $a=1$ gives $x_+=1$ or $x_-=\frac23$; $a=-1$ gives  discriminant of $-23$; and for $|a|\ge2$ we get the discriminant $7-(3a^4+3(2a-1)^2)<0$ (since $3a^4+3(2a-1)^2$ is nonnegative).
Hence $x=1$, $y=0$ is the only solution.
